Let be $E$ a vector space such that $\dim E = 2$.
Let be $G$ a finite subgroup of $GL_2(\mathbb{R})$ such that $G$ generates a subalgebra $\mathcal{A}$ of $\mathcal{L}(E)$ : linear maps from $E$ to $E$ such that $\dim A = 3$.
Show that, for all $f \in G, f^2 = \textrm{Id}$, $\textrm{Id}$ being the identity map.
It is a duplicate of this unanswered question: Order in finite subgroup of GL(E) ; a comment remarks that $\dim (<g_1, \ldots, g_p>) = 2$ where $G = \{ g_1, \ldots, g_p \}$ with $p = \textrm{card } G$ with no proof.
Attempts:
First, as $G$ is a finite subgroup, for all $f \in G$, $f^p = \textrm{Id}$.
So : $\det(f^p) = 1$, i.e. $\det(f)^p = 1$, then as determinant of a real matrix, $\det(f) \in \mathbb{R}$, so that $\det(f) = \pm 1$.
This gives us $f(f - (\textrm{tr} f) f) = \pm \textrm{Id}$ (characteristic polynomial in dimension $2$). Then : $f^{-1} = \pm (f - (\textrm{tr} f) f)$.
What I'd like: to show that $p$ is necessarily even, so that I can have a $+\textrm{Id}$ and $\textrm{tr} f = 2$ so that I can conclude.
Now, I'm stuck.
- $p$ is necessarily even if, for example, I can show that $-\textrm{Id} \in G$, so that $(-\textrm{Id})^p = \textrm{Id}$ is equivalent to $p$ even.
I might have solved it.
Here is an idea:
As $\det f = \pm 1$ and $\dim E = 2$, then $\det f = ab, (a, b) \in \mathbb{R}^2$, so that $(a, b) \in \{ (1, 1), (-1, -1), (1, -1), (-1, 1) \}$. Then, there is three cases which give rise to 3 different trace values, namely: 2, -2, 0. The case 0 is trivial. 2 also, because it gives $f = \textrm{Id}$ and $-2$ is the case when $f = -\textrm{Id}$.
At the end, for all $f \in G, f^2 = \textrm{Id}$.